Curvilinear Motion: Normal and Tangential components (Learn to solve any problem)

Let's go through how to solve Curvilinear motion, normal and tangential components.
More Examples:
• Cars move around the “...
• When the car reaches p...
• The two particles A an...
• The speedboat travels ...
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A toboggan is traveling down along a curve...
The truck travels in a circular path...
The jet plane travels along the vertical parabolic path...
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Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2013.

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

I am really glad it was helpful. Unfortunately, I am unsure when I can upload videos about centroids as I am currently working on thermodynamic videos. Best wishes with your studies!

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

Yes, there are multiple videos covering a lot of the dynamics chapters. Please look through the playlist :)

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[@QuestionSolutions]

You're very welcome! :)

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[@QuestionSolutions]

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[@QuestionSolutions]

Nothing is beyond your capacity of understanding. I am glad it was helpful. Try to do as many questions as you can and you will have an excellent foundation for future topics :)

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[@QuestionSolutions]

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[@cosmosatrosanguineus3171]

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[@QuestionSolutions]

Are you a first year university student? Even if you aren't, it's actually fairly common to feel overwhelmed especially because university courses move so quickly. Create a nice time table with time allocated for each course and take it one step at a time. Seriously, doing things one step at a time is really good for students so they don't get overwhelmed. Also, when you need help, PLEASE use office hours, talk to your professors and TAs. They are usually sitting in an empty office waiting for students to come and ask questions (few comes rarely). Best wishes, and remember, you really can do this!@@cosmosatrosanguineus3171

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[@QuestionSolutions]

I am glad these videos are helpful to you! Best wishes with your studies.

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[@QuestionSolutions]

You're very welcome. Best of luck with your studies.

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[@QuestionSolutions]

I am very happy to hear that and wish you the best with your studies!

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very much helpful sir can you please upload more videos in curvilinear motion advance

[@QuestionSolutions]

Glad it was helpful. There really isn't any more to upload on curvilinear motion, I think there are some solved problems already uploaded as well. I will keep what you said in mind for the future 👍

[@aceheeseung1780]

not me watching this few hours before our exam. thanks a lot!

[@QuestionSolutions]

You got this! Best of luck and you're very welcome.

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Ha! Same.

[@MP-xq2yr]

Thank you so much. Your explanations are crystal clear. Would you please teach us how to find the angle between posetive x-axis and at (tangential acceleration? Thank you in advance. PS. Please upload more videos for dynamic.

[@QuestionSolutions]

Hi, thank you so much for taking the time to write that. I appreciate it, and I am really glad this helped you. To find the angle between the x-axis and the tangential acceleration, please see this example: kzbin.info/www/bejne/m2SkXqybjKubjdE . It covers how to find that. :) Best of luck with your studies!

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[@QuestionSolutions]

You're welcome! Best of luck with your studies.

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[@QuestionSolutions]

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[@keegandevine5949]

In the final plane question, if you convert the X value in the P equation into meters instead of kilometres it gives you an answer of something like 8x10^10m which is obviously way larger than your answer (even after you convert your answer into meters) , so in the calculation even though speed and everything else is in m/s and not km/s for some reason the only correct way to get P is by using km, is there a specific reason for this? previously we could easily work out p by using meters? why is it that this has changed for this specific question? i was always under the impression that you are meant to convert your units within the actual calculation and not once you already have a final answer... could you please assist xd

[@QuestionSolutions]

Great question! There is a small misunderstanding on your part :) The whole P equation is in kilometres, not meters. I think maybe you thought the 0.8 on the bottom is m/s, but that is not it. It is the second derivative of our y=0.4x^2 equation. So to understand, notice that our original equation for the path of the jet is actually an equation given to us in kilometres, not meters. What I mean is, if I substitute 1 into x, we are saying that's 1 km, not 1 m because if you look at the diagram, we see that our distances are given in km. I'll go a bit deeper by giving an example. Let's say I substitute x=2 into the equation. We get y = 1.6 km. That means at x=2 km, y=1.6 km, not 1.6 m. So when we take the derivative of this equation, every equation we get is also in kilometres. This is why you cannot substitute a meter value into the p equation. After everything is done, you can convert your kilometre value into meters. I think it would have been good if the units for the equation are shown, which is usually the case, but since its not, it might be a bit of a pitfall. Regardless, excellent question and I hope this answer helped.

[@ricercook2396]

@@QuestionSolutions thank you! this was quite a nuanced point and I had the same issue. Glad there was a satisfying answer

[@chadbullard979]

If P equation is: y=.4x^2, would y be in km^2? You are squaring x in km to get km^2?@@QuestionSolutions

[@daxthedestroyer7]

You ever thought about coming to ASU and teaching? We could’ve really used you last semester!

[@QuestionSolutions]

😅 Hopefully, anyone anywhere can watch these and gain a good understanding of the concepts, even if I can't be there physically. Many thanks for your comment!

[@munalshrestha188]

Thanks sir! Can you suggest any references book for dynamics

[@QuestionSolutions]

Mechanics for engineers - dynamics by R. C. Hibbeler and K. B. Yap. :)

[@elisabethvanbiesen3166]

I hope you will see this in time, i have a question about the last excercise (with the jet). When you search the radius of curvature, I used as my x value 5000 m instead of using 5 km. This gives a whole other outcome that is impossibly great. Why doesn't it work if you convert the distance to meters. I love your videos btw they help me so much

[@QuestionSolutions]

The equation given is in km. I am really glad these videos helped you out :)

[@ReneZalazar01]

AWESOME VIDEO! Quick question, why is the tangential acceleration equal to the value of the sentence "is increasing at a rate"? Is the tangential acceleration related to the magnitude and the normal acceleration related to the direction?

[@QuestionSolutions]

I guess the easiest answer I can give is this. In almost all the cases, the acceleration you're given in a question is tangential. Normal acceleration is zero unless the object is traveling along a curve. So imagine there is a question and it said a car is accelerating at 2 m/s^2, or the velocity is increasing at a rate of 2 m/s^2. That's tangential acceleration. If the car was on a curve, then you will end up finding normal acceleration. Normal acceleration is related with the radius of curvature and velocity. It will ALWAYS point towards the center of the curve. Tangential acceleration is acceleration along a tangent at an instantaneous point. I hope that helps. Best of luck with your studies :)

[@ReneZalazar01]

@@QuestionSolutions Hell yeah man! Thank you so much! Your videos really helped my out. Keep going please!

[@QuestionSolutions]

@@ReneZalazar01 You're very welcome and I am really glad to hear it helped :)

[@m.zakria8327]

bro is there any chance you graduated somewhere in canada? Because all these examples and the format of the videos matches some of my courses for statics, thermo and dynamics. Which makes it even more helpful for me but I am just curious.

[@QuestionSolutions]

Yes, and glad to hear they are helpful. :)

[@Lexyvil]

Thanks! This helps a lot! However how do I solve a problem like the one at 2:50 when the acceleration is a function of time instead of a function of distance, and that I need to know the velocity at a specific distance without knowing the time it takes?

[@QuestionSolutions]

Glad to hear it helped! It's hard to answer your question without knowing the givens in your questions, but there should be enough values for you to use your kinematics equations. The overall process remains the same.

[@Lexyvil]

@@QuestionSolutions Glad to know, thank you!

[@QuestionSolutions]

@@Lexyvil You're very welcome.

[@romaalexieespeleta9956]

Thank you for this video! But how can you solve a problem requiring the normal acceleration, wherein only the constant speed and diameter of the circle are given?

[@QuestionSolutions]

I show the equation at 0:20. It's velocity divided by the radius. Since you have the diameter, you just divide the diameter by 2 to get the radius. Btw, speed is the velocity (without a direction). 👍

[@romaalexieespeleta9956]

​@@QuestionSolutions I was just hesitant about my solution, but that is what I also did. Thank you so much!

[@QuestionSolutions]

@@romaalexieespeleta9956 You're very welcome! Keep up the good work :)

[@rawans1718]

4:38 1:32 Thanks for helping out I want to ask how I find value of dx

[@QuestionSolutions]

Sorry, but what do you mean by dx? Are you looking for time derivatives?

[@rawans1718]

​@@QuestionSolutionshow find the secound derivative dy/dx=0.02x

[@QuestionSolutions]

So you take the second derivative with respect to x. In this case, the second derivative is just 0.02.@@rawans1718

[@sujalbhatu999]

If we take 5000m instead of 5km in finding radius of curvature than answe comes different , can you please explain

[@QuestionSolutions]

The equation is given in km, not meters.

[@matteobekink9939]

Great video . Could you help me with a similar problem. I need to calculate the angle between the tangential component of the acceleration and the vertical line (y-axis) . Is there a way to relate the y(x) equation such that i can calculate the angle at any given point on the curve?

[@rangertato]

Hey this is the question snipboard.io/aifgKc.jpg

[@QuestionSolutions]

I am not too sure I understand your question, sorry about that. But at 2:13, I show a purple arrow and green arrow. Using those, can you tell me what angle you're trying to find? From what I understand, you want the angle between the green arrow and the y-axis? Are you looking for where it intercepts and the angle there?

[@rabiaalbalushi4344]

On question 2, what equation did you use to relate acceleration to tangential acceleration? When you did a=at=0.05(10).

[@QuestionSolutions]

It's given in the question, so when questions say a car accelerated at 2 m/s^2, that's tangential acceleration. Normal acceleration only occurs when an object is travelling along a curve. So there is nothing to relate accelerating to tangential acceleration since that's what's given in the question.

[@ahmedgri8709]

@@QuestionSolutions very good video, but i still don't understand why would you multiply it by 10, the acceleration should be constant?

[@haikalfauzi_]

for the as ds = v dv used in question 2, why do you subtract 8? where dyou get that?

[@QuestionSolutions]

It's a definite integral, so don't forget to subtract after you plug in the values. (4^2)/2 = 8.

[@johnmichaelgania7700]

does the value of x that is 5km doesn't need to be converted to meters ?

[@QuestionSolutions]

No, since the radius comes out in km, then you can convert to meters.

[@mkhairulirwantarmedi7356]

i love you

[@QuestionSolutions]

❤

[@francisnaong4111]

Thank you for explaining it on a easier manner. BTW, I have a question what if there's a spring connected on the object what should I do ? Thanks

[@QuestionSolutions]

You're very welcome. Hmm, I am not entirely sure I understand the "spring connected" part. Is the object on a curved path? If so, you can use principle of work and energy to figure out acceleration and then it's all the same. Transverse acceleration would be the acceleration you get from your kinematics equations, while normal acceleration can be found using the velocity and radius of curvature. I have not really seen a spring connected problem in this chapter, so I assume you're talking about multiple things happening at once. Regardless of the situation, remember, if it's on a curve, there will be normal acceleration, and the way to find it is with velocity and radius of curvature. If you keep that in mind, the rest is usually trying to find what the velocity is, and what the radius is. I am not sure if this answer helped, but if there is a more specific question, let me know, and I'll do my best. Best of luck with your studies.

[@jadabrown1811]

hello, I am having some confusion with P. Do you know why I might be getting 1202.93?? I'm sure I might be miscalculating something, but I have checked me work so many ways. Let me know if you can figure out my problem. Thank you!! Amazinggg videos!

[@QuestionSolutions]

Please give me a timestamp so I know where you're referring to. That way, I can help you better. Thanks!

[@muhammadsaadnasir2116]

Sir how we're understanding that this velocity and acceleration is normal or tangent?

[@QuestionSolutions]

Acceleration has 2 components, normal and tangential. Tangential acceleration is the stuff you usually talk about. Normal acceleration is acceleration when an object travels through a curve.

[@darrylcarter3691]

The first example I already knew the tangential acceleration was going to be 3m/s^2 because the sled is sliding down the hill therefore it is speeding up. That means it is accelerating. And it is tangential to the curve. Though you said vA dot is 3 m/s^2, we can assume that speed at point A with a dot on top is the same as acceleration. I even know that the radius of Curvature was going to be in use when I saw the y = 0.01x^2. So I didn’t need much help with that one.

[@QuestionSolutions]

Awesome! 😁

[@darrylcarter3691]

@@QuestionSolutions second one I know what had to do and what equations to use. So I did it myself than when I played the video to see how you did it. I realized that I made little booboo. 😅 I just forgot to make v one half v squared. But again I had the right idea. I just needed to slow down a bit. 😅

[@QuestionSolutions]

@@darrylcarter3691 Great! Take it one step at a time, try it out first, and then check and that's the way to go :) Well done!

[@muhammadridhwan1096]

3:57 why we dont use v=4 m/s into the an equation?

[@QuestionSolutions]

We do at 3:45, the lower bound of our integral uses 4 m/s.

[@rubentuting525]

can I see the derivations of these formulas?

[@QuestionSolutions]

You should be able to find all the derivations and proofs of these formulas in your textbook, or any other dynamics book. I don't really want to prove or derive these equations since there are tons of videos done about how these equations come to be. If you would like, I highly suggest Khan academy videos, they show proofs and derivations. I just want to show how to solve problems you face in these chapters, or rather, how to use these equations and when to apply them. Thanks!

[@clearflow7925]

4:48 if i plug in 5000 m instead of 5km, i get a different radius of curvature. why can't i plug it in there so i don't have to convert it back to meters later

[@QuestionSolutions]

The equation is written with respect to km.

[@amirnaidoo2002]

3:43 how did you get the 8 in the equation?

[@QuestionSolutions]

So when you solve the definite integral on the left side, you get v^2/2 - 4^2/2 ==> v^2/2 - 8

[@peachmochi2223]

Hello! I used the formula that you taught in this video. But I'm not sure if I got the correct answer, here is the problem: A particle having a constant speed of 15 ft/s moves around a circle of 10 ft. in diameter. What is the normal acceleration? And my answer is 45 ft/s², may I know if it's correct? Thank you!

[@QuestionSolutions]

Yes, your answer is correct. Normal acceleration is found by dividing velocity squared by the radius. So 15²/5=45 ft/s². 👍

[@peachmochi2223]

@@QuestionSolutions Thank you so much! You really have a good way of teaching.

[@QuestionSolutions]

@@peachmochi2223 Thank you for the kind words. :)

[@saktolang7784]

I was about to cry till I found you

[@QuestionSolutions]

Don't worry, you got this! One step at a time, one question at a time :)

[@howis-goin]

jesus, wish i found you sooner

[@QuestionSolutions]

Better late than never :)

[@think_in_math]

Can I ask you about the name of the program that you presentat with it Your student abdo

[@QuestionSolutions]

The animations are done using after effects and drawings are done on illustrator.

[@yellowdood8445]

how did the radius of curvature in the last example became 87.62? when i plug in the same equation in my calculator it shows 113.05 is this some kind of error? please clarify thanks! 4:50

[@yellowdood8445]

i got it i was wrong with where i put my ^2 thanksss this has been super helpful for practicing questions

[@QuestionSolutions]

Awesome! I am really happy it worked out for you. Keep up the good work.

[@goron3007]

The same thing happened with me ​@@yellowdood8445

[@kabellamajid3950]

The 200m/s if initial velocity and it was increased by certain amount of a=0.8,giving final velocity of 220m/s and its what you use as your V?? Check the 3rd example,please!!!

[@QuestionSolutions]

I don't think you understood the question. At point A, the jet has a velocity of 200 m/s, and we need to find our unknowns at point A. So we use 200 km/s. At that instant, there is an acceleration but at that instant, the speed is only 200 m/s. Imagine you're in a car with an acceleration of 20 m/s^2. Your speedometer shows a speed of 40 km/s. If we are looking for values at that instant, we don't figure out what the speed is going to be at a later stage, we use the value shown on our speedometer. Same thing.

[@Sreevar5454x]

Where r you from sir please love from india AP

[@QuestionSolutions]

Canada 👍

[@JootjeTUDelft]

In the last example when calculating the radius, you use 5 (km) instead of 5000(m). Why shouldnt you use 5000m?

[@user-ed1us9wi9y]

I was also wondering

[@QuestionSolutions]

The equation is given in km.

[@tutstorial8474]

how to get the acc. tangential?

[@QuestionSolutions]

I will do my best to help but I am unsure of what you are asking? Sorry.

[@tutstorial8474]

@@QuestionSolutions the tangential acceleration

[@tutstorial8474]

@@QuestionSolutions or is it always given?

[@QuestionSolutions]

Hmm, from what I remember, it is almost always given. You can always find it using the equations from previous chapters, like a=dv/dt, or vdv=ads, etc.

[@tutstorial8474]

thank you so much, I highly appreciate the time you lend just to reply to my comment sir 😊

[@tetosamuel6294]

4:47 There is a mistake 5000^3

[@QuestionSolutions]

The equation is given with respect to km, if you covert to m, you will get an incorrect answer.

[@RzGarage]

I should get a refund from my school and pay you instead

[@QuestionSolutions]

😅 I am glad it helped you and I wish you the absolute best with your studies. That was a funny comment 🤣